How do I integrate 8x sin (1/6 x) dx?

Answer 1

You need to use integration by parts!

Given

∫8xsin(x/6)dx = 8∫xsin(x/6)dx

Let

u = x

v' = sin(x/6)

u' = 1

v = ∫v'dx = ∫sin(x/6)dx = -6cos(x/6) using integration by substitution

Then

8∫xsin(x/6)dx

= 8[uv - ∫u'v]

= 8[(x)(-6cos(x/6)) - ∫(1)(-6cos(x/6))dx]

= 8[-6xcos(x/6) + 6∫cos(x/6)dx]

= -48cos(x/6) + 48∫cos(x/6)dx

So we now only have one more integral left to solve, and we can use integration by substitution:

48∫cos(x/6)dx = 48 * 6 * sin(x/6) = 288sin(x/6)

Finally, regrouping everything

∫8xsin(x/6)dx = -48cos(x/6) + 288sin(x/6) + C